Dear Roland,
given the integral
(1) Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x] (0<r<a)
Applying your substitution x->r Cos[u] I have found (p=a/r<1)
(2) Integrate[ ArcCosh[ p Sec[u] ],u]
With your previous substitution x-> a/Cosh[u] I have found
(3) -p(NIntegrate[u Tanh[u]/Sqrt[Cosh[u]^2 -p^2],u]
And verified that (1),(2) and (3) give the same results when numerically eva
luated on a finite interval (b,r) with (b<r<a). Unfortunately none of them
seems to be solvable in the realm (rather large!) of the Mathematica8 funct
ions.
May be one has to use higher order trascendent functions? Or, wisely, stop h
ere?
Many thanks again for your attention, Rob
-----Messaggio originale-----
Da: Franzius, R. [mailto:roland.franzius at uos.de]
Inviato: marted=EC 4 dicembre 2012 11.13
A: Brambilla Roberto Luigi (RSE)
Oggetto: Re: R: Re: Difficult antiderivative
Am 30.11.2012 14:45, schrieb Brambilla Roberto Luigi (RSE):
> Dear Roland,
>
> Many thanks for your suggestion, i.e. given the indefinite integral (antid
erivative) with r<a :
>
> 1) Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x]
>
> (that Mathematica8 can't solve) change the variable x->a/Cosh[u].
> Doing the substitution the integral becomes
>
> 2) Integrate[ u Tanh[u]/Sqrt[q^2 Cosh[u]^2-1],x]
>
> where q=r/a (r<a). Unfortunately also this integral is unsolvable by
> Mathematica8 (unless q=1).
>
> Alexei Boulbitch wrote (29 november) [mg128833]
> ....
> most of indefinite integrals have this property ("does not exist") , and o
nly smaller part of them can be expressed in terms of analytical and special
functions.
> ....
> For some integrals you can find the solution, for others you cannot, and n
o general rule exists that would help you to distinguish one group from the
other.
>
> Are (1) and (2) cases of this unhappy class?
>
There was probably a typo somewhere.
I didn't store the procedure and cam't reproduce the complete polylog
results.
The most simple algebraic form may be
replace
Sqrt(r^2-x^2) /. x->r Cos[u]
to get a Sin[u] which is cancled by the derivative
So you are left with
Integrate[ ArcCosh[ p Csc[u] ], u ]
This function is a bit strange, the Argument of ArcCosh has to be >1
demanding
0 < Cos[u] < Min[1,p]
Apply TrigToExp and with some manipulations on the Log you get a term
Integrate[Log[Cos[u]],u ]
an insolvable rest which my look like
Integrate[ Log[1 + Sqrt[ 1- q^2 Cos[u]^2] ], u ]
Regards
Roland Franzius
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